Proving that one solution always lies above the other

[urbanist]

Hi all,

I'd be very happy if you could help me solve a problem in my research.

I need to prove the following:
H'(r) = -y(r) - k H(r)

k is a constant.

y is strictly increasing, but not continuous.

Let (a,b]\subset R.

(H_x, y_x) denotes solution x.

H_1(a)<H_0(a)<0.

H_0(s)<0, H_1(s)<0 for all s\in(a,b].

y_1(s)>y_0(s) for all s \in (a,b].

Show:

H_1(r)<H_0(r) for all r \in (a,b].

 

[pasmith]

Hi all,

I'd be very happy if you could help me solve a problem in my research.

I need to prove the following:
H'(r) = -y(r) - k H(r)

k is a constant.

y is strictly increasing, but not continuous.

Let (a,b]\subset R.

(H_x, y_x) denotes solution x.

H_1(a)<H_0(a)<0.

H_0(s)<0, H_1(s)<0 for all s\in(a,b].

y_1(s)>y_0(s) for all s \in (a,b].

Show:

H_1(r)<H_0(r) for all r \in (a,b].

Starting from
<br /> H_0&#039; + kH_0 = - y_0 \\<br /> H_1&#039; + kH_1 = -y_1<br />
if we subtract the second from the first we obtain
<br /> (H_0 - H_1)&#039; + k(H_0 - H_1) = y_1 - y_0<br />
and if we multiply by e^{kr} we find that
<br /> \frac{d}{dr} \left(e^{kr} (H_0 - H_1) \right) = e^{kr}(y_1 - y_0)<br />
Now the right hand side is strictly positive for all r \in (a,b], and so e^{kr}(H_0 - H_1) is strictly increasing on (a,b]. Since it is initially strictly positive (H_0(a) &gt; H_1(a)), it therefore remains strictly positive. It follows that H_0(r) &gt; H_1(r) for all r \in (a,b] as required.

 

[urbanist]

Thanks a lot! That was a very elegant explanation :)

Now, can I prove the same for:

H&#039;(r)=-y(r)-H(r) k(-H(r)) ,
where k is a strictly increasing positive function?

 

[pasmith]

Thanks a lot! That was a very elegant explanation :)

Now, can I prove the same for:

H&#039;(r)=-y(r)-H(r) k(-H(r)) ,
where k is a strictly increasing positive function?

Yes, assuming that continuous solutions of the now non-linear ODEs exist on [a,b].

Starting from the same point as before, we obtain
<br /> (H_0 - H_1)&#039; + k(-H_0)H_0 - k(-H_1)H_1 = y_1 - y_0<br />
Now k(-H_0)H_0 - k(-H_1)H_1 is not obviously \frac{f&#039;(r)}{f(r)}(H_0 - H_1) for some strictly positive f(r), so we can't use the previous method. However, we can use a different idea.

Suppose there exists r_0 \in (a,b] such that H_0(r_0) = H_1(r_0). It then follows, under the assumptions of the previous problem, that
<br /> (H_0 - H_1)&#039;(r_0) = y_1(r_0) - y_0(r_0) &gt; 0<br />
which means that at that point H_0 - H_1 is strictly increasing. Thus locally H_0(r) &lt; H_1(r) if r &lt; r_0 and H_0(r) &gt; H_1(r) if r &gt; r_0.

It follows that there exists at most one such r_0 \in (a,b], since once H_0 &gt; H_1 the solutions can't intersect again in that interval; if they did then at that point (H_0 - H_1)&#039; would not be strictly positive, which is impossible.

Thus if H_0(a) &gt; H_1(a) then again it must follow that H_0(r) &gt; H_1(r) for all r \in (a,b].

 

[urbanist]

Beautiful solution, once again :)

I was wondering, whether the fact that y is not necessarily continuous matters.

Can I speak about (H_0-H_1)&#039;(r_0) and y_0(r_0)-y_1(r_0)?
Or do I need to deal with (H_0-H_1)&#039;(r\rightarrow r_0^-).
And then, can I claim y_0(\rightarrow r_0^-)-y_1(r\rightarrow r_0^-)&gt;0? I think I can, since y_0(r_0)-y_1(r_0)&gt;0 everywhere... But I'm not sure how to write that formally.

 

[pasmith]

Beautiful solution, once again :)

I was wondering, whether the fact that y is not necessarily continuous matters.

How badly behaved are the functions you are considering? If y is at least piecewise continuous then the above arguments will hold on each subinterval on which y is continuous.

I think there also constraints on y in order for H to exist in the first place.

 

[pasmith]

On closer inspection, it appears that if y_1 - y_0 and k are not continuous then we can't necessarily show that H_0 &gt; H_1.

We want to use the sign of (H_0 - H_1)&#039; to show that if (H_0 - H_1)(r_0) = 0 then H_0 - H_1 is strictly increasing in an open neighborhood of r_0. That is enough for us to conclude that if H_0 - H_1 is initially strictly positive then it remains strictly positive.

More formally, we want to show that there exists \epsilon &gt; 0 such that if |r - r_0| &lt; \epsilon then (H_0 - H_1)&#039;(r) &gt; 0. Then, by the mean value theorem, H_0 - H_1 will be strictly increasing on that interval. It is not necessary for the application of the mean value theorem that (H_0 - H_1)&#039; be continuous. We do however need continuity, or at least suitable limiting behaviour, at r_0 in order to show that a suitable \epsilon &gt; 0 exists.

Now if (H_0 - H_1)&#039; is continuous at r_0 then, since it is strictly positive at r_0, it follows that such an \epsilon exists.

However if (H_0 - H_1)&#039; is not continuous at r_0 but
\lim_{r \to r_0^{+}} (H_0 - H_1)&#039;(r) = \lim_{r \to r_0^{+}} (y_1 - y_0)(r) &gt; 0 and
\lim_{r \to r_0^{-}} (H_0 - H_1)&#039;(r) = \lim_{r \to r_0^{-}} (y_1 - y_0)(r) &gt; 0
then that again guarantees the existence of such an \epsilon (I am assuming here that
<br /> \lim_{r \to r_0} (k(-H_0)H_0 - k(-H_1)H_1)(r) = 0<br />
and if that is not the case then it is not the case that the limits of (H_0 - H_1)&#039; and y_1 - y_0 are equal, and I don't see how to make further progress).

But the most that the condition y_1(r) &gt; y_0(r) gives us is that
\lim_{r \to r_0^{+}} (y_1 - y_0)(r) \geq 0
and
\lim_{r \to r_0^{-}} (y_1 - y_0)(r) \geq 0
if those limits actually exist. If one of those limits is zero then it might be that (H_0 - H_1)&#039; approaches zero only through negative values, and if one of those limits doesn't exist then we can't say anything at all.

 

[urbanist]

Well, we know that y_1(r)&gt;y_0(r) for all r.

If y is defined for all r, isn't it necessarily piecewise continuous?

k, at any rate, is continuous.

It seems to me that there has to be a neighborhood to the left of r_0 in which y_1(r)-y_0(r)&gt;k(-H_0)H_0-k(-H_1)H_1...

 

[pasmith]

Well, we know that y_1(r)&gt;y_0(r) for all r.

If y is defined for all r, isn't it necessarily piecewise continuous?

In general, no. But since y is constrained to be strictly increasing https://www.physicsforums.com/newreply.php?do=newreply&p=4527279 that the only discontinuities it can have are jump discontinuities, and there can only be a countable number of them. It follows that the only discontinuities y_1 - y_0 can have are jump discontinuities, and there can only be a countable number of them. It also follows that the necessary one-sided limits exist.

k, at any rate, is continuous.

Excellent; my approach should work.

It seems to me that there has to be a neighborhood to the left of r_0 in which y_1(r)-y_0(r)&gt;k(-H_0)H_0-k(-H_1)H_1...

I'm not convinced that this is necessarily the case.